Bài tập 3.58 trang 184 SBT Toán 12

Ta có: \(\displaystyle  \tan x = 0 \Leftrightarrow x = 0\) do \(\displaystyle  x \in \left[ { - \frac{\pi }{4};\frac{\pi }{4}} \right]\).

Khi đó \(\displaystyle  S = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left| {\tan x} \right|dx} \) \(\displaystyle   = \int\limits_{ - \frac{\pi }{4}}^0 {\left| {\tan x} \right|dx}  + \int\limits_0^{\frac{\pi }{4}} {\left| {\tan x} \right|dx} \) \(\displaystyle   =  - \int\limits_{ - \frac{\pi }{4}}^0 {\tan xdx}  + \int\limits_0^{\frac{\pi }{4}} {\tan xdx} \)

\(\displaystyle   =  - \int\limits_{ - \frac{\pi }{4}}^0 {\frac{{\sin x}}{{\cos x}}dx}  + \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x}}{{\cos x}}dx} \) \(\displaystyle   = \int\limits_{ - \frac{\pi }{4}}^0 {\frac{{d\left( {\cos x} \right)}}{{\cos x}}}  - \int\limits_0^{\frac{\pi }{4}} {\frac{{d\left( {\cos x} \right)}}{{\cos x}}dx} \) \(\displaystyle   = \left. {\ln \left| {\cos x} \right|} \right|_{ - \frac{\pi }{4}}^0 - \left. {\ln \left| {\cos x} \right|} \right|_0^{\frac{\pi }{4}}\)

\(\displaystyle   = \ln 1 - \ln \frac{{\sqrt 2 }}{2} - \ln \frac{{\sqrt 2 }}{2} + \ln 1\) \(\displaystyle   =  - 2\ln \frac{{\sqrt 2 }}{2} = \ln 2\)

Chọn C.

-- Mod Toán 12 HỌC247